leetcode101- Symmetric Tree- easy

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1
   / 
  2   2
 /  / 
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1
   / 
  2   2
      
   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

1. 递归：这题关键是把内部镜像转化为外部互动镜像，这样才能递归下去，否则怎么都不能把总树自成镜像问题拆成子树自成镜像问题啊，丢失了信息。

一个树是镜像的就是它的两个子树互成镜像。两棵树互成镜像就是两棵树的根相等且树1.左和树2.右互成镜像，且树1.右和树2.左互成镜像。

2.迭代：类似于BFS（不需要层级遍历）。就是每次加入的点的顺序要按镜像来。

细节：一开始可以传入两次root来解决第一个点略不一样的问题

1.递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        return helper(root, root);
    }
    
    private boolean helper(TreeNode n1, TreeNode n2) {
        
        if (n1 == null && n2 == null) {
            return true;
        }
        if (n1 == null || n2 == null) {
            return false;            
        }
        return n1.val == n2.val && helper(n1.left, n2.right) && helper(n1.right, n2.left);
    }
}

2. 迭代

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {

        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
            // 注意等于与否不要担心null，只要不一样就会输出false. null == null-> true, null != null -> false
            // null == 某个体 -> false
            // 记得要比value，不是比个体(地址)
            if (n1 == null && n2 == null) {
                continue;
            }
            if (n1 == null || n2 == null || n1.val != n2.val) {
                return false;
            }
            queue.offer(n1.left);
            queue.offer(n2.right);
            queue.offer(n1.right);
            queue.offer(n2.left);
            
        }
        return true;
    }
}

3.自己写的很丑的迭代

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {

        Queue<TreeNode> queue = new LinkedList<TreeNode>();

        if (root == null || root.left == null && root.right == null) {
            return true;
        } else if (root.left == null || root.right == null || root.left.val != root.right.val) {
            return false;            
        } else {
            queue.offer(root.left);
            queue.offer(root.right);
        }
        
        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i + 1 < size; i += 2) {
                TreeNode n1 = queue.poll();
                TreeNode n2 = queue.poll();
                // 注意等于与否不要担心null，只要不一样就会输出false. null == null-> true, null != null -> false
                // null == 某个体 -> false
                // 记得要比value，不是比个体(地址)
                if(n1.left == null && n2.right != null || n1.left != null && n2.right == null ||
                   n2.left == null && n1.right != null || n2.left != null && n1.right == null) {
                    return false;                    
                }
                if (n1.left != null && n2.right != null && n1.left.val != n2.right.val) {
                    return false;
                }
                if (n1.right != null && n2.left != null && n1.right.val != n2.left.val) {
                    return false;
                }
                if (n1.left != null) {
                    queue.offer(n1.left);
                    queue.offer(n2.right);
                }
                if (n1.right != null) {
                    queue.offer(n1.right);
                    queue.offer(n2.left);
                }
            }
        }
        return true;
    }
}